\newproblem{lay:7_3_13}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.3.13}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ be a symmetric $n\times n$ matrix, let $M$ and $m$ denote the maximum and minimum values of the quadratic form $\mathbf{x}^TA\mathbf{x}$, and denote corresponding
	unit eigenvectors by $\mathbf{u}_1$ and $\mathbf{u}_n$. The following calculations show that given any number $t$ between $M$ and $m$, there is a unit vector
	$\mathbf{x}$ such that $t=\mathbf{x}^TA\mathbf{x}$. Verify that $t=(1-\alpha)m+\alpha M$ for some number $\alpha$ between 0 and 1. Then, let
	$\mathbf{x}=\sqrt{1-\alpha}\mathbf{u}_n+\sqrt{\alpha}\mathbf{u}_1$, and show that $\mathbf{x}^T\mathbf{x}=1$ and $\mathbf{x}^TA\mathbf{x}=t$.
}{
   % Solution
	Let us first show that any number $t$ between $m$ and $M$ can be written as
	\begin{center}
		$t=(1-\alpha)m+\alpha M$
	\end{center}
	with $\alpha\in[0,1]$. If $\alpha=0$, we get $t=m$. If $\alpha=1$, we get $t=M$. We see that $t=(1-\alpha)m+\alpha M=m+\alpha(M-m)$ is a linear (and, therefore, continuous)
	function of $\alpha$. So for any value $t$ between 0 and 1, there exists a value of $\alpha$ such that $t=(1-\alpha)m+\alpha M$.
	
	Now, we construct
	\begin{center}
		$\mathbf{x}=\sqrt{1-\alpha}\mathbf{u}_n+\sqrt{\alpha}\mathbf{u}_1$
	\end{center}
	Let's check that $\mathbf{x}^T\mathbf{x}=1$, for which we will exploit the fact that $\mathbf{u}_1$ and $\mathbf{u}_n$ are unitary and orthogonal
	\begin{center}
		$\begin{array}{rcl}
		   \mathbf{x}^T\mathbf{x}&=&(\sqrt{1-\alpha}\mathbf{u}_n+\sqrt{\alpha}\mathbf{u}_1)^T(\sqrt{1-\alpha}\mathbf{u}_n+\sqrt{\alpha}\mathbf{u}_1)\\
			    &=&(\sqrt{1-\alpha}\mathbf{u}_n^T+\sqrt{\alpha}\mathbf{u}_1^T)(\sqrt{1-\alpha}\mathbf{u}_n+\sqrt{\alpha}\mathbf{u}_1)\\
			    &=&(1-\alpha)\mathbf{u}_n^T\mathbf{u}_n+\alpha\mathbf{u}_1^T\mathbf{u}_1+\sqrt{1-\alpha}\sqrt{\alpha}\mathbf{u}_n^T\mathbf{u}_1+
					    \sqrt{1-\alpha}\sqrt{\alpha}\mathbf{u}_1^T\mathbf{u}_n\\
					&=&(1-\alpha)+\alpha+0+0\\
					&=&1
		\end{array}$
	\end{center}
	
	Finally, we need to show that $\mathbf{x}^TA\mathbf{x}=t$, remind that $\mathbf{u}_1$ is the eigenvector associated to the eigenvalue $M$ and that
	$\mathbf{u}_n$ is the eigenvector associated to eigenvalue $m$:
	\begin{center}
		$\begin{array}{rcl}
		   \mathbf{x}^TA\mathbf{x}&=&(\sqrt{1-\alpha}\mathbf{u}_n+\sqrt{\alpha}\mathbf{u}_1)^TA(\sqrt{1-\alpha}\mathbf{u}_n+\sqrt{\alpha}\mathbf{u}_1)\\
			    &=&(\sqrt{1-\alpha}\mathbf{u}_n^T+\sqrt{\alpha}\mathbf{u}_1^T)(\sqrt{1-\alpha}A\mathbf{u}_n+\sqrt{\alpha}A\mathbf{u}_1)\\
			    &=&(\sqrt{1-\alpha}\mathbf{u}_n^T+\sqrt{\alpha}\mathbf{u}_1^T)(m\sqrt{1-\alpha}\mathbf{u}_n+M\sqrt{\alpha}\mathbf{u}_1)\\
			    &=&(m(1-\alpha)\mathbf{u}_n^T\mathbf{u}_n+M\alpha\mathbf{u}_1^T\mathbf{u}_1+M\sqrt{1-\alpha}\sqrt{\alpha}\mathbf{u}_n^T\mathbf{u}_1+
					    m\sqrt{1-\alpha}\sqrt{\alpha}\mathbf{u}_1^T\mathbf{u}_n\\
			    &=&m(1-\alpha)+M\alpha+0+0\\
			    &=&t\\
		\end{array}$
	\end{center}
}
\useproblem{lay:7_3_13}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

